Limit Problem Solution Overview

This problem asks us to find the value of a specific limit as \(x\) approaches 1. The function is given by the expression \(\mathop {\lim }\limits_{x \to 1} \left( {\frac{{1 - {e^{ - c\left( {1 - x} \right)}}}}{{1 - {xe^{ - c\left( {1 - x} \right)}}}}} \right)\). To solve this limit, we first need to check the form of the expression when \(x=1\).

Checking the Indeterminate Form

Let's substitute \(x=1\) into the numerator and the denominator of the given expression to identify its form:

• Numerator Evaluation: When \(x = 1\), the numerator becomes \(1 - {e^{ - c\left( {1 - 1} \right)}} = 1 - {e^{ - c\left( 0 \right)}} = 1 - {e^0} = 1 - 1 = 0\).
• Denominator Evaluation: When \(x = 1\), the denominator becomes \(1 - {1 \cdot e^{ - c\left( {1 - 1} \right)}} = 1 - {e^{ - c\left( 0 \right)}} = 1 - {e^0} = 1 - 1 = 0\).

Since both the numerator and the denominator evaluate to 0 at \(x=1\), the limit is of the indeterminate form \(\frac{0}{0}\). This condition allows us to apply L'Hopital's Rule to determine the limit's value.

Applying L'Hopital's Rule for Limit Evaluation

L'Hopital's Rule is a powerful technique in calculus that states if \(\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}}\) results in an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then the limit can be found by evaluating \(\mathop {\lim }\limits_{x \to a} \frac{{f'(x)}}{{g'(x)}}\), provided this latter limit exists.

Let's define our functions \(f(x)\) and \(g(x)\) from the given limit expression:

• Let \(f(x) = 1 - {e^{ - c\left( {1 - x} \right)}}\).
• Let \(g(x) = 1 - {xe^{ - c\left( {1 - x} \right)}}\).

Calculating the Derivative of the Numerator, \(f'(x)\)

We differentiate \(f(x)\) with respect to \(x\):

\(f'(x) = \frac{d}{dx}\left( {1 - {e^{ - c\left( {1 - x} \right)}}} \right)\) Using the chain rule, for the term \(- {e^{ - c\left( {1 - x} \right)}}\), let \(u = -c(1-x) = -c + cx\). Then \(\frac{du}{dx} = c\). So, \(\frac{d}{dx}(e^u) = e^u \cdot \frac{du}{dx}\).

Therefore, \(f'(x) = 0 - \left( {{e^{ - c\left( {1 - x} \right)}} \cdot c} \right)\) \(f'(x) = - c{e^{ - c\left( {1 - x} \right)}}\).

Calculating the Derivative of the Denominator, \(g'(x)\)

We differentiate \(g(x)\) with respect to \(x\). For the term \(xe^{ - c\left( {1 - x} \right)}\), we need to apply the product rule \(\frac{d}{dx}(uv) = u'v + uv'\), where \(u = x\) and \(v = {e^{ - c\left( {1 - x} \right)}}\).

• Derivative of \(u = x\) is \(u' = 1\).
• Derivative of \(v = {e^{ - c\left( {1 - x} \right)}}\) is \(v' = {e^{ - c\left( {1 - x} \right)}} \cdot \frac{d}{dx}\left( { - c\left( {1 - x} \right)} \right) = {e^{ - c\left( {1 - x} \right)}} \cdot c\).

Applying the product rule to \(xe^{ - c\left( {1 - x} \right)}\): \(\frac{d}{dx}\left( {{xe^{ - c\left( {1 - x} \right)}}} \right) = \left( 1 \right){e^{ - c\left( {1 - x} \right)}} + x\left( {c{e^{ - c\left( {1 - x} \right)}}} \right)\) \(= {e^{ - c\left( {1 - x} \right)}} + xc{e^{ - c\left( {1 - x} \right)}}\) \(= {e^{ - c\left( {1 - x} \right)}}\left( {1 + xc} \right)\).

Now, substitute this back into \(g'(x)\): \(g'(x) = \frac{d}{dx}\left( {1 - {xe^{ - c\left( {1 - x} \right)}}} \right) = 0 - \left( {{e^{ - c\left( {1 - x} \right)}}\left( {1 + xc} \right)} \right)\) \(g'(x) = - {e^{ - c\left( {1 - x} \right)}}\left( {1 + xc} \right)\).

Evaluating the Limit of the Derivatives Ratio

Now, we apply L'Hopital's Rule by taking the limit of the ratio of \(f'(x)\) and \(g'(x)\):

\(\mathop {\lim }\limits_{x \to 1} \frac{{f'(x)}}{{g'(x)}} = \mathop {\lim }\limits_{x \to 1} \frac{{ - c{e^{ - c\left( {1 - x} \right)}}}}{{ - {e^{ - c\left( {1 - x} \right)}}\left( {1 + xc} \right)}}\)

We can cancel the common term \( - {e^{ - c\left( {1 - x} \right)}}\) from both the numerator and the denominator, as exponential functions are always positive and never zero.

The expression simplifies to: \(\mathop {\lim }\limits_{x \to 1} \frac{c}{{1 + xc}}\)

Finally, substitute \(x = 1\) into this simplified expression to get the limit's value:

Value of the limit = \(\frac{c}{{1 + 1 \cdot c}} = \frac{c}{{1 + c}}\).

Final Limit Value

The value of the limit \(\mathop {\lim }\limits_{x \to 1} \left( {\frac{{1 - {e^{ - c\left( {1 - x} \right)}}}}{{1 - {xe^{ - c\left( {1 - x} \right)}}}}} \right)\) is \(\frac{c}{{c + 1}}\).